a jar contains $8.75 in nickels, dimes and quarters. There are two times as many dimes as there are nickels, and 10 more dimes than quarters, how many of each coin are in the jar?
Let x = number of dimes, then x/2 for nickels and
x-10 for quarters.
5x/2 + 10x + 25(x-10) = 875
Solve for x, then x/2 and x-10.
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To solve this problem, we can set up a system of equations based on the given information.
Let's assume that the number of nickels in the jar is "n", the number of dimes is "d", and the number of quarters is "q".
From the given information, we know the following:
1. The value of nickels is 5 cents, so the total value of nickels can be calculated as: 0.05n.
2. The value of dimes is 10 cents, so the total value of dimes can be calculated as: 0.10d.
3. The value of quarters is 25 cents, so the total value of quarters can be calculated as: 0.25q.
We are also given the following conditions:
1. The total value of all coins in the jar is $8.75, so we can write the equation: 0.05n + 0.10d + 0.25q = 8.75.
2. There are two times as many dimes as there are nickels, so we can write the equation: d = 2n.
3. There are 10 more dimes than quarters, so we can write the equation: d = q + 10.
Now we have a system of three equations with three variables. We can solve this system using substitution or elimination to find the values of n, d, and q.
From equation 2, we can substitute d = 2n into equation 3:
2n = q + 10.
Now, from equation 1, we substitute d = 2n and q = 2n - 10:
0.05n + 0.10(2n) + 0.25(2n - 10) = 8.75.
Simplifying:
0.05n + 0.20n + 0.50n - 2.50 = 8.75.
Combining like terms:
0.75n - 2.50 = 8.75.
Adding 2.50 to both sides:
0.75n = 11.25.
Dividing both sides by 0.75:
n = 15.
Now, we can use this value of n to find d and q.
From equation 2, since d = 2n:
d = 2(15) = 30.
From equation 3, since d = q + 10:
30 = q + 10.
Subtracting 10 from both sides:
q = 20.
Therefore, there are 15 nickels, 30 dimes, and 20 quarters in the jar.